Precalculus Examples

$\frac{{(y + 2)}^{2}}{16} - \frac{{(x - 3)}^{2}}{9} = 1$

Step 1

Simplify each term in the equation in order to set the right side equal to $1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be $1$ .

$\frac{{(y + 2)}^{2}}{16} - \frac{{(x - 3)}^{2}}{9} = 1$

Step 2

This is the form of a hyperbola. Use this form to determine the values used to find vertices and asymptotes of the hyperbola.

$\frac{{(y - k)}^{2}}{a^{2}} - \frac{{(x - h)}^{2}}{b^{2}} = 1$

Step 3

Match the values in this hyperbola to those of the standard form. The variable $h$ represents the x-offset from the origin, $k$ represents the y-offset from origin, $a$ .

$a = 4$

$b = 3$

$k = - 2$

$h = 3$

Step 4

The center of a hyperbola follows the form of $(h, k)$ . Substitute in the values of $h$ and $k$ .

$(3, - 2)$

Step 5

Find $c$ , the distance from the center to a focus.

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Step 5.1

Find the distance from the center to a focus of the hyperbola by using the following formula.

$\sqrt{a^{2} + b^{2}}$

Step 5.2

Substitute the values of $a$ and $b$ in the formula.

$\sqrt{{(4)}^{2} + {(3)}^{2}}$

Step 5.3

Simplify.

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Step 5.3.1

Raise $4$ to the power of $2$ .

$\sqrt{16 + {(3)}^{2}}$

Step 5.3.2

Raise $3$ to the power of $2$ .

$\sqrt{16 + 9}$

Step 5.3.3

Add $16$ and $9$ .

$\sqrt{25}$

Step 5.3.4

Rewrite $25$ as $5^{2}$ .

$\sqrt{5^{2}}$

Step 5.3.5

Pull terms out from under the radical, assuming positive real numbers.

$5$

Step 6

Find the vertices.

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Step 6.1

The first vertex of a hyperbola can be found by adding $a$ to $k$ .

$(h, k + a)$

Step 6.2

Substitute the known values of $h$ , $a$ , and $k$ into the formula and simplify.

$(3, 2)$

Step 6.3

The second vertex of a hyperbola can be found by subtracting $a$ from $k$ .

$(h, k - a)$

Step 6.4

Substitute the known values of $h$ , $a$ , and $k$ into the formula and simplify.

$(3, - 6)$

Step 6.5

The vertices of a hyperbola follow the form of $(h, k \pm a)$ . Hyperbolas have two vertices.

$(3, 2), (3, - 6)$

Step 7

Find the foci.

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Step 7.1

The first focus of a hyperbola can be found by adding $c$ to $k$ .

$(h, k + c)$

Step 7.2

Substitute the known values of $h$ , $c$ , and $k$ into the formula and simplify.

$(3, 3)$

Step 7.3

The second focus of a hyperbola can be found by subtracting $c$ from $k$ .

$(h, k - c)$

Step 7.4

Substitute the known values of $h$ , $c$ , and $k$ into the formula and simplify.

$(3, - 7)$

Step 7.5

The foci of a hyperbola follow the form of $(h, k \pm \sqrt{a^{2} + b^{2}})$ . Hyperbolas have two foci.

$(3, 3), (3, - 7)$

Step 8

Find the eccentricity.

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Step 8.1

Find the eccentricity by using the following formula.

$\frac{\sqrt{a^{2} + b^{2}}}{a}$

Step 8.2

Substitute the values of $a$ and $b$ into the formula.

$\frac{\sqrt{{(4)}^{2} + {(3)}^{2}}}{4}$

Step 8.3

Simplify the numerator.

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Step 8.3.1

Raise $4$ to the power of $2$ .

$\frac{\sqrt{16 + 3^{2}}}{4}$

Step 8.3.2

Raise $3$ to the power of $2$ .

$\frac{\sqrt{16 + 9}}{4}$

Step 8.3.3

Add $16$ and $9$ .

$\frac{\sqrt{25}}{4}$

Step 8.3.4

Rewrite $25$ as $5^{2}$ .

$\frac{\sqrt{5^{2}}}{4}$

Step 8.3.5

Pull terms out from under the radical, assuming positive real numbers.

$\frac{5}{4}$

Step 9

Find the focal parameter.

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Step 9.1

Find the value of the focal parameter of the hyperbola by using the following formula.

$\frac{b^{2}}{\sqrt{a^{2} + b^{2}}}$

Step 9.2

Substitute the values of $b$ and $\sqrt{a^{2} + b^{2}}$ in the formula.

$\frac{3^{2}}{5}$

Step 9.3

Raise $3$ to the power of $2$ .

$\frac{9}{5}$

Step 10

The asymptotes follow the form $y = \pm \frac{a (x - h)}{b} + k$ because this hyperbola opens up and down.

$y = \pm \frac{4}{3} \cdot (x - (3)) - 2$

Step 11

Simplify to find the first asymptote.

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Step 11.1

Remove parentheses.

$y = \frac{4}{3} \cdot (x - (3)) - 2$

Step 11.2

Simplify $\frac{4}{3} \cdot (x - (3)) - 2$ .

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Step 11.2.1

Simplify each term.

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Step 11.2.1.1

Multiply $- 1$ by $3$ .

$y = \frac{4}{3} \cdot (x - 3) - 2$

Step 11.2.1.2

Apply the distributive property.

$y = \frac{4}{3} x + \frac{4}{3} \cdot - 3 - 2$

Step 11.2.1.3

Combine $\frac{4}{3}$ and $x$ .

$y = \frac{4 x}{3} + \frac{4}{3} \cdot - 3 - 2$

Step 11.2.1.4

Cancel the common factor of $3$ .

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Step 11.2.1.4.1

Factor $3$ out of $- 3$ .

$y = \frac{4 x}{3} + \frac{4}{3} \cdot (3 (- 1)) - 2$

Step 11.2.1.4.2

Cancel the common factor.

$y = \frac{4 x}{3} + \frac{4}{3} \cdot (3 \cdot - 1) - 2$

Step 11.2.1.4.3

Rewrite the expression.

$y = \frac{4 x}{3} + 4 \cdot - 1 - 2$

Step 11.2.1.5

Multiply $4$ by $- 1$ .

$y = \frac{4 x}{3} - 4 - 2$

Step 11.2.2

Subtract $2$ from $- 4$ .

$y = \frac{4 x}{3} - 6$

Step 12

Simplify to find the second asymptote.

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Step 12.1

Remove parentheses.

$y = - \frac{4}{3} \cdot (x - (3)) - 2$

Step 12.2

Simplify $- \frac{4}{3} \cdot (x - (3)) - 2$ .

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Step 12.2.1

Simplify each term.

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Step 12.2.1.1

Multiply $- 1$ by $3$ .

$y = - \frac{4}{3} \cdot (x - 3) - 2$

Step 12.2.1.2

Apply the distributive property.

$y = - \frac{4}{3} x - \frac{4}{3} \cdot - 3 - 2$

Step 12.2.1.3

Combine $x$ and $\frac{4}{3}$ .

$y = - \frac{x \cdot 4}{3} - \frac{4}{3} \cdot - 3 - 2$

Step 12.2.1.4

Cancel the common factor of $3$ .

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Step 12.2.1.4.1

Move the leading negative in $- \frac{4}{3}$ into the numerator.

$y = - \frac{x \cdot 4}{3} + \frac{- 4}{3} \cdot - 3 - 2$

Step 12.2.1.4.2

Factor $3$ out of $- 3$ .

$y = - \frac{x \cdot 4}{3} + \frac{- 4}{3} \cdot (3 (- 1)) - 2$

Step 12.2.1.4.3

Cancel the common factor.

$y = - \frac{x \cdot 4}{3} + \frac{- 4}{3} \cdot (3 \cdot - 1) - 2$

Step 12.2.1.4.4

Rewrite the expression.

$y = - \frac{x \cdot 4}{3} - 4 \cdot - 1 - 2$

Step 12.2.1.5

Multiply $- 4$ by $- 1$ .

$y = - \frac{x \cdot 4}{3} + 4 - 2$

Step 12.2.1.6

Move $4$ to the left of $x$ .

$y = - \frac{4 x}{3} + 4 - 2$

Step 12.2.2

Subtract $2$ from $4$ .

$y = - \frac{4 x}{3} + 2$

Step 13

This hyperbola has two asymptotes.

$y = \frac{4 x}{3} - 6, y = - \frac{4 x}{3} + 2$

Step 14

These values represent the important values for graphing and analyzing a hyperbola.

Center: $(3, - 2)$

Vertices: $(3, 2), (3, - 6)$

Foci: $(3, 3), (3, - 7)$

Eccentricity: $\frac{5}{4}$

Focal Parameter: $\frac{9}{5}$

Asymptotes: $y = \frac{4 x}{3} - 6$ , $y = - \frac{4 x}{3} + 2$

Step 15